/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 A wall's height is two-thirds th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A wall's height is two-thirds that of its length. The entire wall will be painted. The paint costs \(\$ 12\) per gallon, and 1 gallon of paint covers 60 square feet. What expression gives the cost of the paint (assuming one can purchase partial and full gallons) to cover a wall that is \(\mathrm{L}\) feet long? (A) \(\quad\) Cost \(=L^{2}\) (B) \(\quad\) Cost \(=\frac{1}{5} L^{2}\) (C) \(\quad\) Cost \(=\frac{2}{15} L^{2}\) (D) \(\quad\) Cost \(=\frac{3}{64} L^{2}\)

Short Answer

Expert verified
Option (C): \( \frac{2}{15}L^2 \).

Step by step solution

01

Determine the height of the wall

The height of the wall is described as two-thirds of its length. If the wall's length is \( L \), the height is \( \frac{2}{3}L \).
02

Calculate the area of the wall

The area of a rectangular wall is given by the formula \( \, \text{Area} = \text{length} \times \text{height} \, \). Therefore, the area of the wall is \( L \times \frac{2}{3}L = \frac{2}{3}L^2 \).
03

Determine gallons of paint needed

Since 1 gallon of paint covers 60 square feet, the number of gallons needed is the total area divided by coverage per gallon, i.e., \( \frac{\frac{2}{3}L^2}{60} \). Simplifying this gives \( \frac{L^2}{90} \).
04

Calculate the cost of the paint

The cost of the paint is the number of gallons needed multiplied by the cost per gallon. The expression for the cost is \( \frac{L^2}{90} \times 12 = \frac{L^2}{7.5} \), which simplifies to \( \frac{2}{15}L^2 \).
05

Identify the correct answer choice

Option (C) \( \frac{2}{15}L^2 \) matches our derived expression for the cost of the paint.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometry
Geometry plays a crucial role in understanding the spatial aspects of different objects, such as a wall in this exercise. Geometry helps us comprehend dimensions, shapes, and the area of constructs we encounter in mathematical problems.
When determining the dimensions of the wall, we start with its length, denoted as \( L \). The height of the wall is then described as \( \frac{2}{3}L \), showing a clear relationship between length and height. This proportionality is a fundamental concept in geometry. It illustrates how one dimension can influence another, which is a common occurrence in real-life scenarios as well.
In this exercise, understanding this concept allows us to calculate the wall's area, a key step in determining the cost of materials needed. To find the area, we use the formula for the area of a rectangle: length times height. Therefore, the area of the wall is given by \( L \times \frac{2}{3}L \), which computes to \( \frac{2}{3}L^2 \).
Recognizing this geometric relationship enables us to solve problems related to coverage and cost, forming a basis for further algebraic calculations.
Mathematical Expressions
Mathematical expressions are equations or formulas that represent relationships or operations on quantities. They allow us to model and solve real-world problems efficiently.
In this problem, the mathematical expression for the area \( \frac{2}{3}L^2 \) is derived from geometric measurements. This expression helps in the subsequent calculations, such as determining the amount of paint needed. Each part of the expression holds a meaning—a reflection of physical dimensions transcribed into mathematical language.
Using expressions simplifies complex scenarios, translating them into manageable parts. When considering the paint coverage, we formulated \( \frac{\frac{2}{3}L^2}{60} \) as an expression to calculate the gallons needed. This expression represents the relationship between the area to be painted and the coverage per gallon. Simplifying this expression leads us to \( \frac{L^2}{90} \), making it easier to handle computationally.
Mathematical expressions provide clarity and precision, essential for understanding and solving problems systematically.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying expressions to solve equations or make calculations simpler and clearer. This process is a vital skill in handling mathematical problems efficiently.
In this exercise, after determining the expression for the number of gallons of paint, \( \frac{L^2}{90} \), we move on to calculating the cost. The manipulation involves multiplying this expression by the cost per gallon, \( 12 \), yielding \( \frac{L^2}{90} \times 12 \).
Careful simplification is crucial as it reduces the expression to \( \frac{L^2}{7.5} \), which can be further simplified to \( \frac{2}{15}L^2 \). This final expression represents the total cost of the paint in a concise and clear form.
By algebraically manipulating expressions, we attain results that are not only precise but also optimized for problem-solving. These skills are indispensable for efficiently progressing through mathematical exercises and real-world applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following could be a value of \(x\) in this equation? $$ 8 x^{2}=-16 x-2 $$ I. \(-1-\frac{\sqrt{3}}{2}\) II. \(\frac{1}{2}(-2-\sqrt{6})\) III. \(\frac{1}{2}(\sqrt{3}-2)\) (A) I only (B) II only (C) \(\mid\) and IIl only (D) \(\quad\) II and III only

The amount of money \((A)\) in a bank account after a principal amount \((P)\) is on deposit for \(t\) years at an annual interest rate \(r\) compounded \(n\) times per year is given by this equation: $$ A=P\left(1+\frac{r}{n}\right)^{n t} $$ Suppose that a banker would like to determine how changes in these variables would cause the bank to pay less interest to its clients. Which of the variables \(-P, r, n\), and \(t-\) if minimized, would cause less interest paid to clients? (A) \(P\) only (B) \(r\) and \(t\) only (C) \(n\) and \(t\) only (D) \(P, r, n\), and \(t\)

Given that \(i=\sqrt{-1}\), what is the value of \(i^{4}+i^{12}\) ? (A) \(\sqrt{-1}\) (B) \(-1\) (C) 1 (D) 2

First solve the problem, and then enter your answer on the grid provided on the answer sheet. The instructions for entering your answers follow. The variables \(m\) and \(n\) have a directly proportional relationship given by the equation \(m=k n\), where \(k\) is a constant of proportionality. When \(m=10, n=2\). What will be the value of \(n\) if \(m\) equals \(38 ?\)

Which of the following options for the underlined portion would best elaborate on the topic of the sentence with the most specific detail? (A) NO CHANGE (B) were constituted of a great deal of mass and matter, and possessed sufficient gravitational pull to have a significant impact on their surroundings, (C) had enough mass to become nearly round due to pressure, and had enough gravity to clear their orbit of any other bodies, (D) had a tremendously large amount of mass, while enough gravity to be quite noticeable,

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.